Question: A company has $500$ employees, and $60\%$ of them have children. Suppose that we randomly select $4$ of these employees. Which of the following would find the probability that exactly $3$ of the $4$ employees selected have children? Choose 1 answer: Choose 1 answer: (Choice A) A $(0.60)^3(0.40)$ (Choice B) B $(0.60)(0.40)^3$ (Choice C) C ${4 \choose 3}(0.60)^3(0.40)$ (Choice D) D ${4 \choose 3}(0.60)(0.40)^3$ (Choice E) E ${500 \choose 4}(0.60)^3(0.40)$
Probability of $3$ successes We want the probability that there are $3$ successes (employee has children) in $4$ trials (employees), so we're going to need $1$ failure (employee doesn't have children) as well. The probability of each success is ${60\%}$ and the probability of each failure is $40\%}$. We can assume independence since we are sampling less than $10\%$ of the population, so we can multiply probabilities to find the probability of getting $3$ successes followed by $1$ failure: $\begin{aligned} P(\text{SSSF})&=({0.60})({0.60})({0.60})(0.40}) \\\\ &=({0.60})^3(0.40}) \end{aligned}$ The binomial coefficient ${n \choose k}$ SSSF isn't the only arrangement that produces $3$ successes in $4$ trials. For instance, FSSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=3$ successes (employee has children) in $n=4$ trials (employees), so we should use the binomial coefficient ${4 \choose 3}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $(0.60)^3(0.40)$ so for our final answer we multiply this probability by the number of possible arrangements: ${4 \choose 3}(0.60)^3(0.40)$ The answer: ${4 \choose 3}(0.60)^3(0.40)$